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python - 'int' 对象没有属性 'append'

转载 作者:太空宇宙 更新时间:2023-11-04 09:13:43 24 4
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当我运行此代码时遇到以下错误,我是编程新手,我知道我有一堆无用的数组。我不知道我的错误在哪里,因为我已经将 j 声明为数组。我完全没有想法。

import pyodbc,nltk,array,re,itertools
cnxn = pyodbc.connect('Driver={MySQL ODBC 5.1 Driver};Server=127.0.0.1;Port=3306;Database=information_schema;User=root; Password=1234;Option=3;')
cursor = cnxn.cursor()
cursor.execute("use collegedatabase ;")
cursor.execute("select * from sampledata ; ")
cnxn.commit()
s=[]
j=[]
x=[]
words = []
w = []
sfq = []
POS=[]
wnl = nltk.WordNetLemmatizer()
p = []
clean= []
l =[]
tupletolist= []
results = []
aux = []
regex = re.compile("\w+\.")
pp = []
array1=[]

f = open("C:\\Users\\vchauhan\\Desktop\\tupletolist.txt","w")
for entry in cursor:
s.append(entry.injury_type),j.append(entry.injury_desc)

def isAcceptableChar(character):
return character not in "~!@#$%^&*()_+`1234567890-={}|:<>?[]\;',/."


from nltk.tokenize import word_tokenize
from nltk.corpus import stopwords
english_stops = set(stopwords.words('english'))
for i in range(0,200):
j.append(filter(isAcceptableChar, j[i]))
w.append([word for word in word_tokenize(j[i].lower()) if word not in english_stops])
for j in range (0,len(w[i])):
results = regex.search(w[i][j])
if results:
str.rstrip(w[i][j],'.')
for a in range(0 , 200):
sfq.append(" ".join(w[a]))

from nltk.stem import LancasterStemmer
stemmer = LancasterStemmer()

for i in range (0,200):
pp.append(len(w[i]))

for a in range (0,200):
p.append(word_tokenize(sfq[a]))
POS.append([wnl.lemmatize(t) for t in p[a]])
x.append(nltk.pos_tag(POS[a]))
clean.append((re.sub('()[\]{}'':/\-[(",)]','',str(x[a]))))
cursor.execute("update sampledata SET POS = ? where SRNO = ?", (re.sub('()[\]{}'':/\-[(",)]','',str(x[a]))), a)

for i in range (0,len(array1)):
results.append(regex.search(array1[i][0]))
if results[i] is not None:
aux.append(i)

f.write(str(w))

异常(exception):

Traceback (most recent call last):
File "C:\Users\vchauhan\Desktop\regexsolution_try.py", line 37, in <module>
j.append(filter(isAcceptableChar, j[i]))
AttributeError: 'int' object has no attribute 'append'

最佳答案

j 已被用作列表和整数。仅将 j 用作整数名称,将列表命名为其他名称。

j.append(filter(isAcceptableChar, j[i]))    # j is not a list here,it is an int.
w.append([word for word in word_tokenize(j[i].lower()) if word not in english_stops])
for j in range (0,len(w[i])): # here j is an int

关于python - 'int' 对象没有属性 'append',我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11877665/

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